We can further generalize our previous problem and replace Δ by any elliptic self-adjoint operator. We will soon specialize to the case of the Laplace–Beltrami operator, but this section is completely general.
Our original motivation was to understand the index of operators, so let us start from there. Suppose we are given a compact Riemannian manifold M and E,F Hermitian vector bundles on M. We are also given an elliptic differential operator D:Γ(E)→Γ(F) over M with formal adjoint D∗:Γ(F)→Γ(E) (in the case of the Laplace–Beltrami operator, we have D=D∗=d+d∗ and E=F=Ω∗).
Similar to the case of the Laplace–Beltrami operator, we define
This is the form of the index that will be of interest to us.
By analogy with the classical heat equation, we consider the equation
(∂t∂+ΔE)f=0
on E×(0,∞)→M×(0,∞), with t∈(0,∞).
Hodge theory gives us an easy way to solve this, at least formally. We can decompose
L2(E)=λ⨁Γλ(E),
where Γλ(E) is the λ-eigenspace of ΔE. The spectral theorem tells us the eigenvalues are discrete and tend to infinity. Let {ψλ} be an orthonormal eigenbasis. Then we can write a general solution as
f(x,t)=λ∑cλe−λtψλ(x)
for some constants cλ.
If we want to solve this with initial condition f0, i.e. we require f(−,t)→f0 as t→0 in L2, then we must pick1
f(x,t)=λ∑e−λtψλ(x)⟨ψλ,f0⟩.
Thus, we can write the heat kernel as
Ht(x,y;ΔE)=λ∑eλtψλ(x)ψλ(y)T.
Then we have
f(x,t)=∫MHt(x,y;ΔE)f0(y)dy.
We remark that for each fixed t, the heat kernel Ht(x,y) is a section of the exterior product E⊠E∗→M×M.
To ensure the sum converges, we need to make sure the eigenvalues grow sufficiently quickly. This is effectively Weyl's law, but we for our purposes, we can use a neat trick to obtain a weaker bound easily.
Lemma1.1
If Δ is any self-adjoint elliptic differential operator, then there is a constant C and an exponent ε such that for large Λ, the number of eigenvalues of magnitude ≤Λ is at most CΛε.
Proof
□
To simplify notation, we assume Δ acts on the trivial line bundle. By replacing Δ with its powers, we may assume that the order d of Δ is large enough such that we can apply the Sobolev (and regularity) bound
∥f∥C0≤C′∥f∥d≤C(∥Δf∥L2+∥f∥L2).
In particular, if ψ is an eigenfunction of eigenvalue at most Λ, then we can bound
∥ψ∥C0≤C(1+Λ)∥ψ∥L2.
Let {ψλ} be an orthonormal eigenbasis. Then for any constants aλ and fixed x∈M, we have
∣∣∣λ∣≤Λ∑cλψλ(x)∣∣≤C(1+Λ)(λ≤Λ∑∣cλ∣2)1/2.
We now pick cλ to be the numbers ψˉλ(x), recalling that we have fixed x. Then
Integrating over all of M, the left-hand side is just the number of eigenvalues of magnitude ≤Λ. So we are done.
Proof
□
Elliptic regularity lets us bound the higher Sobolev norms of Ht as well, and so we know Ht is in fact smooth by Sobolev embedding.
It is fruitful to consider the “time evolution” operator e−tΔE that sends f0 to f(−,t) defined above. This acts on the λ eigenspace by multiplication by e−λt. Thus, the trace is given by
ht(ΔE)≡tre−tΔE=λ∑e−λtdimΓλ(E).
The key observation is that D gives an isomorphism between the λ eigenspace of ΔE and the λ eigenspace of ΔF as long as λ>0, and the 0 eigenspaces are exactly the kernels of ΔE and ΔF. So we have an expression
indexΔE=ht(ΔE)−ht(ΔF)
for any t>0.
This expression is a very global one, because ht(ΔE) depends on the eigenfunctions of ΔE. However, the fact that e−λΔE is given by convolution with Ht(x,y;ΔE) gives us an alternative expression for the trace, namely
ht(ΔE)=∫MHt(x,x;ΔE)dx.
This is still non-local, but we will later find that the asymptotic behaviour of Ht(x,x;ΔE) as t→0 is governed by local invariants. Since the index is independent of t, we can take the limit t→0 and get a local expression for the index.
Example1.2
Take the flat torus Td=Rd/Zd, and pick D=d:Ω0(M)→Ω1(M), so that Δ is the classical Laplacian
Δ=−∑∂xi2∂2.
The eigenvectors are given by e2πix⋅ξ for ξ∈Zd with eigenvalue 4π2∣ξ∣2. So the heat kernel is